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UCC2809D-2 Design, How to decrease the Start-up Time ???

Rick Bogdan
Intellectual 930 points
Other Parts Discussed in Thread: UCC2809-2, UCC2809-1

Attached is a one page schematic of a UCC2809D-2 Power supply.  The supply receives +40 volts DC and produces +7.5 volts DC. Everything work fine, except it takes approximately 145 msec to come up. We need it to be in the 20 msec or less range. I checked all the on-line docs for the UCC2809 but couldn't find any help calibrating the Start-up time .. Any suggestion appreciated ...

  • 0
    TI__Expert 4375 points

    Rick,

     

    How are you measuring start up time? is it from applying VIN to when VOUT is in regulation or is it when VOUT rises from zero to 7.5V?

     

    If it is the former (which is what I believe) then start up time is caused by your start up circuit, R400, C403 and C404. Doing the math on these I get a start up time of ~132ms. Reducing R400 to 3k will reduce this time to ~20ms, but power dissipation in R400 will increase to ~0.3W. Lower VIN's will increase your start up time.

    I do notice that the input range is 10V to 40V, per the schematic. The UCC2809-2 needs a minmum of 10.4V to start so this range will not work. If the input is 11V and R400=20k it will take over a second to start up. With 3k this drops to 165ms.

     

    To get fast start up across the input range you could replace R400 with a current source this will also reduce power loss

     

    Richard.

  • 0 in reply to Richard Garvey
    TI__Expert 4375 points

    Actually the UCC2809-2 needs 15.6V to start, the above math would be correct for the UCC2809-1.

     

    For the UCC2809-2 the start up time with R400=20k would be ~200ms. to get the 20ms R400 would have to be <2k

     

    Richard